Explanation

(d)
$PbCl_{2} \rightleftharpoons {Pb}^{2 +} + 2I^{-}$
$s$ mol/L $s$ $2s$
Hence, $K_{sp} = \left\lbrack {Pb}^{2 +} \right\rbrack\left\lbrack I^{-} \right\rbrack^{2}$
$= (s)(2s)^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4s^{3}$
$\therefore$ $s = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{2.3 \times 10^{- 32}}{10}}$
$= 1.78 \times 10^{- 11}$mol/L
$= 1.78 \times 10^{- 11} \times 278\$g/L
$= 4.95 \times 10^{- 9}$ g/L